Evaluation of△U,AHof Ideal-GasExampleThe air is heated during constant pressure flow processAnd its flux is qm = 0.5kg/s .The inlet temperature is 300K andoutlet temperature is 400K.Determine the heat flux to air16
16 Evaluation of of Ideal U,H -Gas Example The air is heated during constant pressure flow process. And its flux is .The inlet temperature is 300K and outlet temperature is 400K. Determine the heat flux to air. 0.5kg/s m q =
Solutionp=constantAccording to 1st LawWdp + AHQ=W +△H=-Q=△H1. Constant Specific HeatTab.3Cp = 1.004kJ/(kgK)Q= △H= qmC,(T, -T)=0.5×1.004 ×(400- 300)=50.2 kJ/s17
17 Solution According to 1st Law d 2 t 1 Q W H V p H Q H = + = − + = p=constant 1. Constant Specific Heat Tab.3 2 1 1.004kJ/(kg K) ( ) =0.5 1.004 (400 300)=50.2 kJ/s p m p c Q H q c T T = = = − −
Tab.22. Empirical EquationIc,dT = Im(a +a,T+a,T? +a,T3)dTAH=qMmaa2 T3αi T223M4= 50.65kJ/sa = 28.15a, =1.967×10-3az = 4.801x10-6as = -1.966x10-918
18 2. Empirical Equation Tab.2 2 2 2 3 0 1 2 3 1 1 2 1 2 2 3 4 3 0 1 d ( )d ( ) 234 50.65kJ/s m m p m q H q c T a a T a T a T T M q a a a a T T T T M = = + + + = + + + = 0 3 1 6 2 9 3 28.15 1.967 10 4.801 10 1.966 10 a a a a − − − = = = = −
3.Average Specific HeatTab.4&5t =27℃, t, =127℃200Pl。=1.004kJ/(kg·K),c,l=1.006kJ/(kg.K),c,l。=1.012kJ/(kg·K)采用插值法27127=1.00454kJ/(kg·K),cplo=1.00762kJ/(kg ·K)CpO= qm△h = qm(c,"-t2 -c,let.= 0.5×(1.00762×127-1.00454×27)=50.4kJ/s4. Thermodynamic Property Table of Ideal-GasTab.6~12Q= qm(h400k -ho0k)= 0.5x(400.98-300.19) = 50.4kJ/s19
19 3. Average Specific Heat Tab.4 & 5 1 2 0 100 200 0 0 0 27 127 1.004kJ/(kg K), 1.006kJ/(kg K), 1.012kJ/(kg K) p p p t t c c c = = = = = ℃， ℃ 4. Thermodynamic Property Table of Ideal-Gas Tab.6～12 400 300 ( ) 0.5 (400.98 300.19) 50.4kJ/s Q q h h = − m K K = − = 采用插值法 27 127 0 0 1.00454kJ/(kg K), 1.00762kJ/(kg K) p p c c = = 127 27 2 1 0 0 ( ) 0.5 (1.00762 127 1.00454 27) 50.4kJ/s Q q h q c t c t = = − m m p p = − =
Entropy of Ideal-GasSq? for reversible process ds duTpdydsTTSq = du + pdy?1st Law(s pv= RT,du =c,dT? for Ideal-GasdTdv+RdsOTV20
20 Entropy of Ideal-Gas • for reversible process δ d q s T = •1 st Law δq u p v = + d d • for Ideal-Gas ,d dv pv RT u c T = = d d d v T v s c R T v = + d d d u p s v T T = +